In the following reaction, oxygen is the excess reactant. sicl4 + o2 → sio2 + cl2 the table shows an experimental record for the above reaction. experimental record trial starting amount of sicl4 starting amount of o2 actual yield of sio2 1 120 g 240 g 38.2 g 2 75 g 50 g 25.2 g calculate the percentage yield for sio2 for trial 1. also, determine the leftover reactant for the trial. show your work. based on the percentage yield in trial 2, explain what ratio of reactants is more efficient for the given reaction.

              (1)  %age Yield  =  90.03 %

              (2)  Leftover O₂  =  217.4 g

              (3)  Following Ratios of Reactants for efficient reactions:

Case 1:

                                            SiCl₄   :   O₂

                                            75 g    :   14.12 g

                                       0.44 mol  :  0.44 mol

Case 2:

                                            SiCl₄   :   O₂

                                         265 g    :   50 g

                                       1.56 mol  :  1.56 mol

The balance chemical equation is as follow;

                                    SiCl₄ + O₂  →  SiO₂ + 2 Cl₂

Given Data For Trial 1:

Starting Amount of SiCl₄  =  120 g

Starting Amount of O₂      =  240 g

Actual Yield of SiO₂          =  38.2 g

Calculating the percentage yield for SiO₂ for Trial 1;

According to balance chemical equation 1 mole (169.89 g) SiCl₄ will completely react with 1 mole (32 g) O₂ to produce SiO₂ and Cl₂. Therefore, when 120 g (0.70 mole) of SiO₄ will react with 22.60 g (0.70 mole) of O₂. But , we are provided with 240 g of O₂ therefore, O₂ is in excess and the final yield will be controlled by the amount of limiting SiO₄.

So,

As,

          169.89 g (1 mole) SiO₄ produced  =  60.08 g (1 mole) SiO₂

So,

              120 g (0.70 mole) SiO₂ will produce  =  X g of SiO₂

Solving for X,

                     X =  60.08 g × 120 g ÷ 169.89 g

                    X  =  42.43 g of SiO₂

Hene, the theoretical yield is 42.43 g of SiO₂.

Percentage yiels is given as,

                        %age yield  =  Actual Yield / Theoretical Yield × 100

Putting values,

                        %age Yield  =  38.2 g / 42.43 g × 100

                        %age Yield  =  90.03 %

Determining the leftover reactant for the trial 1;

As shown above, O₂ is in excess. For given amount of SiCl₄ only 22.60 g of O₂ is consumed. As we are given with 240 g of O₂ so, the leftover O₂ is calculated as.

                           Leftover O₂  =  Given O₂  -  Used O₂

                           Leftover O₂  =  240 g  -  22.60 g

                           Leftover O₂  =  217.4 g

Given Data For Trial 2:

Starting Amount of SiCl₄  =  75 g

Starting Amount of O₂      =  50 g

Actual Yield of SiO₂          =  25.2 g

Calculating ratio of reactants more efficient for the given reaction;

The efficient reaction is said to be that reaction in which all the reactants are consumed (almost completely) without leaving any leftovers as that we have seen in case of O₂ in trial 1.

According to balance equation the more efficient molar ratio of SiCl₄ and O₂ is,

                                                  1  :  1

Means, for each mole of either SiCl₄ or O₂ we exactly require 1 mole of O₂ or SiCl₄. In Trial 2 we are given with 75 g (0.44 mole) of SiCl₄ and 50 g (1.56 mole) of O₂. Hence, to attain maximum efficience we will require either of the two ratios of reactants.

Case 1:

                                            SiCl₄   :   O₂

                                            75 g    :   14.12 g

                                       0.44 mol  :  0.44 mol

Case 2:

                                            SiCl₄   :   O₂

                                         265 g    :   50 g

                                       1.56 mol  :  1.56 mol

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

indium, and thallium have three electrons i think